A fair coin is tossed four times. What is the probability that the sequence of tosses is HHHT? ?

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  • hace 1 mes

    P(HHHT)=[(1/2)^3](1/2)=(1/2)^4=1/16.

  • hace 1 mes

    None of the answers so far, allow for unconventional results. Such as settling on edge or not landing at all. Oh - and they ignore schrodinger principle, where the coin is captured in flight and exists in an indeterminate state. But......this is why I got into trouble in high school math, for being a smartass. I wasn't - these are conditions a good software designer accounts for. Hey - I never even mentioned Heiseberg !!!

  • hace 1 mes

    1 in 16.  there are 16 possible orders that can arise.  that is one of the 16.  2x2x2x2.

  • hace 1 mes

    p(HHHT) =

    1/2^4 =

    1/16

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  • hace 1 mes

    There are 16 possible sequences, each with the same probability. So the probability of any particular sequence is 1 / 16.

  • hace 1 mes

    There are 2 possibilities that can occur at each toss.

    so, 2 x 2 x 2 x 2 = 16 ways in total

    H, H, H and T is a specific order and one unique combination

    i.e. 1/16

    :)>

  • !!!
    Lv 7
    hace 1 mes

    1/2x1/2x1/2x1/2=1/16.

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